Hello,
I need to know (connected closed) Lie subgroups of SO(2)×So(n), indeed these are compact Lie subgroups of SO(2,n) which I am looking for. But I don't know what we can say about Lie subgroups of product Lie groups. I'll be thankful if someone help me in this subject.
Thanks in advanced.

2 Answers
2

It is not clear to me what kind of answer is expected. Generally speaking, subgroups of Lie groups can be classified by Lie correspondence combined with combinatorial analysis resulting from structure theory of semisimple Lie groups. Below I address two particular cases.

If $K$ is a compact connected semisimple subgroup of $SO(2)\times SO(n)$ then its projection onto the first factor is trivial and the question is reduced to the $SO(n)$ case. (Closed) Lie subgroups of $SO(n)$ are precisely (compact) Lie groups with a faithful $n$-dimensional real orthogonal representation, so there are quite a few of them (the maximal connected ones were classified long time ago by Dynkin). If you need a complete description for small values of $n$, the Atlas of Lie groups is very handy.

In the other extreme case where $K=SO(2)$ you are, in effect, asking about the maps

$$f: SO(2)\to SO(2)\times SO(n).$$

They can be classified by passing to the Lie algebras. More precisely, the differential of $f$ is a linear map $so(2)\to so(2)\oplus so(n).$ Identifying $so(2)$ with $\mathbb{R}$ and $so(n)$ with the skew-symmetric matrices, it may be viewed as a pair $(d,A),$ where $d$ is an integer and $A$ is a skew-symmetric matrix whose eigenvalues are integral multiples of $i.$ Explicitly,

is the counterclockwise rotation by $\varphi$ and $\exp$ is the matrix exponential function.

The maps $f$ and $f'$ associated with non-zero pairs $(d,A)$ and $(d',A')$ have the same image if and only if the pairs are proportional. The case $d=0$ corresponds to an $SO(2)$ subgroup of the second factor $SO(n).$ In the case $d=1$, the subgroup $f(K)$ is the graph of a map $SO(2)\to SO(n).$

Note that for the original question about subgroups of $SO(2,n)$ one must impose further equivalences in the case $n=2$, because different subgroups of $SO(2)\times SO(2)$ can be conjugate in $SO(2,2)$.

I wasn't able to make bmatrix to work correctly. Feel free to fix it if you can.
–
Victor ProtsakSep 2 '12 at 0:45

I fixed your rotation matrix. You need to triple up the backslashes (use \\\ instead of \\ for reasons of how the website and the LaTeX interact here) or else use \cr instead of \\.
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Theo BuehlerSep 2 '12 at 2:33

thanks for your useful comments. I need to know the compact subgroups up to conjugacy, but if it is not possible to do, we may add some nice hypothesis like as semisimle , etc. It is also helpful to know any special classes of such subgroups.
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nerd-mathSep 2 '12 at 14:27

I don't understand, why in the first case (second paragraph of the answer) when $K$ is semisimple it's projection on the first factor is trivial?
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nerd-mathSep 20 '12 at 9:41

The first factor is abelian and semisimple groups do not have non-trivial characters.
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Victor ProtsakSep 20 '12 at 19:49

Let me answer this question for the Lie algebras, which already goes part way to answering the original question. The question is then what are the Lie subalgebras of $\mathfrak{so}(2)\oplus\mathfrak{so}(n)$. The general question of which are the Lie subalgebras of the direct sum of Lie algebras is solved by (the Lie algebraic version of) the Goursat Lemma.

It does not hurt to work in more generality. So Let $\mathfrak{g}_L$ and $\mathfrak{g}_R$ be two real Lie algebras and let $\mathfrak{g} = \mathfrak{g}_L \oplus \mathfrak{g}_R$ be their product. Elements of $\mathfrak{g}$ are pairs $(X_L,X_R)$ with $X_L \in \mathfrak{g}_L$ and $X_R \in \mathfrak{g}_R$. The Lie bracket in $\mathfrak{g}$ of two such elements $(X_L,X_R)$ and $(Y_L,Y_R)$ is given by the pair $([X_L,Y_L], [X_R,Y_R])$.

We are interested in Lie subalgebras $\mathfrak{h}$ of $\mathfrak{g}$.

Let $\pi_L : \mathfrak{g} \to \mathfrak{g}_L$ and $\pi_R : \mathfrak{g} \to \mathfrak{g}_R$ denote the projections onto each factor: they are Lie algebra homomorphisms. Let $\mathfrak{h}_L$ and $\mathfrak{h}_R$ denote, respectively, the image of the subalgebra $\mathfrak{h}$ under $\pi_L$ and $\pi_R$. They are Lie subalgebras of $\mathfrak{g}_L$ and $\mathfrak{g}_R$, respectively. Let us define $\mathfrak{h}^0_L := \pi_L(\ker \pi_R \cap \mathfrak{h})$ and $\mathfrak{h}^0_R := \pi_R(\ker \pi_L \cap \mathfrak{h})$. One checks that they are ideals of $\mathfrak{h}_L$ and $\mathfrak{h}_R$, respectively. This means that on $\mathfrak{h}_L/\mathfrak{h}^0_L$ and $\mathfrak{h}_R/\mathfrak{h}^0_R$ we can define Lie algebra structures. Goursat's Lemma says that these two Lie algebras are isomorphic.

Goursat's Lemma suggests a systematic approach to the determination of the Lie subalgebras of $\mathfrak{g}_L \oplus \mathfrak{g}_R$, which is particularly feasible when $\mathfrak{g}_L$ and $\mathfrak{g}_R$ have low dimension.

Namely, we look for Lie subalgebras $\mathfrak{h}_L \subset \mathfrak{g}_L$ and $\mathfrak{h}_R \subset \mathfrak{g}_R$ which have quotients isomorphic to $\mathfrak{q}$, say. Let $f_L:\mathfrak{h}_L \to \mathfrak{q}$ and $f_R:\mathfrak{h}_R \to \mathfrak{q}$ be the corresponding surjections. Let $\varphi$ denote an automorphism of $\mathfrak{q}$. Then we may define a Lie subalgebra $\mathfrak{h}$ of $\mathfrak{h}_L \oplus \mathfrak{h}_R$ by

A commonly occurring special case is when one of $\mathfrak{h}_L \to \mathfrak{q}$ or $\mathfrak{h}_R \to \mathfrak{q}$ is an isomorphism. For definiteness let us assume that it is $\mathfrak{h}_R \to \mathfrak{q}$ which is an isomorphism. Then we get a Lie algebra homomorphism $\mathfrak{h}_L \to \mathfrak{h}_R$ obtained by composing $\mathfrak{h}_L \to \mathfrak{q}$ with the inverse of $\mathfrak{h}_R \to \mathfrak{q}$. In fact, we get a family of such homomorphisms labelled by the automorphisms of $\mathfrak{q}$ or, equivalently, of $\mathfrak{h}_R$. The fibred product which Goursat's Lemma describes is now the graph in $\mathfrak{h}_L \oplus \mathfrak{h}_R$ of such a homomorphism $\mathfrak{h}_L \to \mathfrak{h}_R$. The resulting Lie algebra is abstractly isomorphic to $\mathfrak{h}_L$.

In your case, since $\mathfrak{h}_R$, say, is one-dimensional, $\mathfrak{q}$ is either trivial or else isomorphic to $\mathfrak{h}_R$. In the trivial case, you have direct products of subalgebras, and in the latter case, the previous paragraph applies.

The Lie subalgebras of $\mathfrak{so}(n)$ are tabulated at least for small $n$ in several places. If you are a physicist, then perhaps Slansky's Physics Report Group theory for unified model building might be the most readable.